# physic

A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2.000 s.) The length of a seconds pendulum is 0.9928 m at Tokyo and 0.9943 m at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?
(g Cambridge / g Tokyo)?
any help?

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asked by sisca
1. T=2PI(L/g)

so the ratio is ...

1=sqrt(.9928/gtokio*gengland/.9943)

squaring both sides..

gengland/gtokyo= .9943/.9928

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2. awesome, Thanks a lot...

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posted by sisca

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